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Differentiation — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentiation5 Marks
Question
If $\cos\text{y}=\text{x}\cos(\text{a}+\text{y}),$ where $\cos\text{a}\neq\pm1,$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{\cos^2(\text{a}+\text{y})}{\sin\text{a}}$
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Answer

consider the given function,
$\cos\text{y}=\text{x}\cos(\text{a}+\text{y}),$ where $\cos\text{a}\neq\pm1$
Differentiating both sides w.r.t. 'x' we get
$-\sin\text{y}\frac{\text{dy}}{\text{dx}}=\text{x}\Big(-\sin(\text{a}+\text{y})\frac{\text{dy}}{\text{dx}}\Big)+\cos(\text{a}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}\big[\text{x}\sin(\text{a}+\text{y})-\sin\text{y}\big]=\cos(\text{a}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{\cos(\text{a}+\text{y})}{\text{x}\sin(\text{a}+\text{y})-\sin\text{y}}$
Multiplying the numerator and the denominator
by $\cos(\text{a}+\text{y})$ on the R.H.S., we have,
$\frac{\text{dy}}{\text{dx}}=\frac{\cos^2(\text{a}+\text{y})}{\text{x}\cos(\text{a}+\text{y})\sin(\text{a}+\text{y})-\cos(\text{a}+\text{y})\sin\text{y}}$
$=\frac{\cos^2(\text{a}+\text{y})}{\text{x}\cos(\text{a}+\text{y})\sin(\text{a}+\text{y})-\cos(\text{a}+\text{y})\sin\text{y}}$
$\big[\because\cos\text{y}=\text{x}\cos(\text{a}+\text{y}),\text{given function}\big]$
$=\frac{\cos^2(\text{a}+\text{y})}{\sin\big[(\text{a}+\text{y})-\text{y}\big]}=\frac{\cos^2(\text{a}+\text{y})}{\sin\text{a}}$

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